Question

[didudsui]

Cho cac so nguyen a, b, c. d thoa man \(a^3+b^3=2\left(c^3-8d^3\right)\). Chung minh rang \(a+b+c+d⋮3\)

Answer 1

[Kiệt Nguyễn]

Ta có: \(a^3+b^3=2\left(c^3-8d^3\right)\)

\(\Leftrightarrow a^3+b^3=2c^3-16d^3\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3c^3-15d^3=3\left(c^3-5d^3\right)\)

\(VP⋮3\Rightarrow a^3+b^3+c^3+d^3⋮3\)(1)

Ta có: \(a^3-a+b^3-b+c^3-c+d^3-d\)

\(=\left(a-1\right)a\left(a+1\right)+\left(b-1\right)b\left(b+1\right)\)

\(+\left(c-1\right)c\left(c+1\right)+\left(d-1\right)d\left(d+1\right)\)

Vì tích 3 số tự nhiên liên tiếp chia hết cho 3 nên \(\left(a-1\right)a\left(a+1\right)+\left(b-1\right)b\left(b+1\right)\)

\(+\left(c-1\right)c\left(c+1\right)+\left(d-1\right)d\left(d+1\right)\)chia hết cho 3 (2)

Từ (1) và (2) suy ra \(a+b+c+d⋮3\left(đpcm\right)\)